Science:Math Exam Resources/Courses/MATH102/December 2017/Question 10 (d)
• Q1 • Q2 • Q3 • Q4 • Q5 • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 • Q9 (a) • Q9 (b) • Q9 (c) • Q10 (a) • Q10 (b) • Q10 (c) • Q10 (d) • Q11 (a) • Q11 (b) • Q12 (a) • Q12 (b) • Q12 (c) • Q12 (d) • Q13 (a) • Q13 (b) • Q13 (c) • Q13 (d) • Q14 (a) • Q14 (b) • Q15 (a) • Q15 (b) • Q16 •
Question 10 (d) |
---|
An animal foraging in a food patch has a net energy gain of by the time it has been foraging for min (where is nonnegative and ). (d) Indicate the optimal time (time to maximize , ) in each case directly on the graph. Justify your answers with a sentence explaining the connection to the calculation in part (b) of this problem. |
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! |
Hint |
---|
By part (b), we need to find a secant line from the origin to the graph so that the slope is as large as possible. |
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
|
Solution |
---|
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. By the Hint, for Patch 1 a secant line has maximum slope just a bit before reaches its maximum. An approximate optimal time is . For Patch 2, the slope is largest when . Hence the optimal time is . To summarize, the optimal time (indicated by the red dots) for Patches 1 and 2 are as follows. |